# 2007 Alabama ARML TST Problems/Problem 14

## Problem

Find the sum of the real roots of

$$x^6+145x^4-2007x^3+145x^2+1=0.$$

## Solution

### Step 1: using the symmetry

Let $p(x)=x^6+145x^4-2007x^3+145x^2+1$

The polynomial $p$ is symmetric. Each symmetric polynomial has the following property: for $x\not=0$, $x$ is a root if and only if $\frac 1x$ is.

To see why this is true, assume that we have a $x\not=0$ such that $p(x)=x^6+145x^4-2007x^3+145x^2+1=0$. We can now divide both sides by $x^6$. We get:

$$1 + \frac{145}{x^2} - \frac{2007}{x^3} + \frac{145}{x^4} + \frac{1}{x^6} = 0$$

But the left hand side is just $p\left(\frac 1x\right)$. Thus also $p\left(\frac 1x\right)=0$.

### Step 2: multiplying three quadratic terms

Our polynomial has six complex roots: $\alpha$, $1/\alpha$, $\beta$, $1/\beta$, $\gamma$, and $1/\gamma$. It can be expressed as:

$$p(x) = (x-\alpha)(x-1/\alpha)(x-\beta)(x-1/\beta)(x-\gamma)(x-1/\gamma)$$

Let $a=\alpha+1/\alpha$, $b=\beta+1/\beta$, and $c=\gamma+1/\gamma$. We can then write:

$$p(x) = (x^2 - ax + 1)(x^2 - bx + 1)(x^2 - cx + 1)$$

We can now compare the coefficients of $x$, $x^2$ and $x^3$ in this expression. (Comparing other coefficients is not necessary due to symmetry.) We get the following equations:

• $-a-b-c=0$
• $3+ab+ac+bc=145$
• $-abc-2a-2b-2c=-2007$

Simplifying, we get:

• $a+b+c=0$
• $ab+ac+bc=142$
• $abc=2007$

Using Vieta's formulas, this means that $a$, $b$ and $c$ are the roots of the polynomial $q(x)=x^3 - 0x^2 + 142x - 2007$.

### Step 3: Finding a root of q

Obviously, $q$ is increasing for $x>0$, and $q(0)<0$. Thus $q$ has exactly one positive root. We can easily find that $q(9)=0$. Thus WLOG $a=9$.

(Note: Remembering that $a+b+c=0$ and $abc=2007$, we now have $b+c=-a=-9$ and $bc=2007/9=223$. From the first equation, let $c=-9-b$. Substituting into the second one, we get $b(b+9)=-223$. This obviously has no real solution, thus $b$ and $c$ are not real. But we don't need this observation.)

### Step 4: Partially dividing p

We now know that $(x^2-9x+1)$ is a factor of $p$. We can partially divide $p$ to get:

$$p(x) = (x^2 - 9x + 1)(x^4 + 9x^3 + 225x^2 + 9x + 1)$$

### Step 5: Handling the remaining part of p

The polynomial $r(x)=x^4 + 9x^3 + 225x^2 + 9x + 1$ has no real roots.

To prove this, one can for example write:

$$r(x) = \left(x^4 + 9x^3 + \frac{81}{4}\cdot x^2\right) + \left(\frac{81}{4}\cdot x^2 + 9x + 1\right) + \left(225-2\cdot\frac{81}{4}\right) x^2$$

$$r(x) = x^2\left(x+\frac92\right)^2 + \left(\frac{9x}2+1\right)^2 + \frac{369}2\cdot x^2$$

### Step 6: Summary

We found that $p(x)=(x^2 - 9x + 1)r(x)$, where $r$ has no real roots. Thus all real roots of $p$ are the two real roots of $x^2 - 9x + 1$, and their sum is obviously $\boxed{9}$.

### Note

  roots of p computed by Octave:
-4.48006 + 14.30270i
-4.48006 - 14.30270i
8.88748 +  0.00000i
0.11252 +  0.00000i
-0.01994 +  0.06367i
-0.01994 -  0.06367i


## See also

 2007 Alabama ARML TST (Problems) Preceded by:Problem 13 Followed by:Problem 15 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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